Question

我构建了一个grep命令，用于递归搜索文件目录中的模式。问题是grep只返回模式所在的文件名，而不是模式的完全匹配。如何返回实际结果？

示例：

文件somefile.bin在包含一百万个其他文件的目录中包含somestring0987654321�123�45�

命令：

$ grep -EsniR -A 1 -B 1 '([a-zA-Z0-9]+)\x00([0-9]+)\x00([0-9]+)\x00' *

目前的结果：

Binary file somefile.bin matches

期望的结果（或接近它）：

Binary file somefile.bin matches

<line above match>
somestring0987654321�123�45�
<line below match>

Answer 1

您可以尝试-a选项：

File and Directory Selection
   -a, --text
          Process  a binary file as if it were text; this is equivalent to
          the --binary-files=text option.

   --binary-files=TYPE
          If the first few bytes of a file indicate that the file contains
          binary  data, assume that the file is of type TYPE.  By default,
          TYPE is binary, and grep  normally  outputs  either  a  one-line
          message  saying  that  a  binary  file matches, or no message if
          there is no match.  If TYPE is without-match, grep assumes  that
          a  binary  file  does  not  match;  this is equivalent to the -I
          option.  If TYPE is text, grep processes a binary file as if  it
          were  text;  this is equivalent to the -a option.  Warning: grep
          --binary-files=text might output binary garbage, which can  have
          nasty  side  effects  if  the  output  is  a terminal and if the
          terminal driver interprets some of it as commands.

但问题是在二进制文件中没有行，所以我不确定你想要的输出是什么样的。你会看到随机垃圾，也许是整个文件，可能会打印一些与你的终端混乱的特殊字符。

如果要将输出限制为匹配本身，请考虑-o选项：

   -o, --only-matching
          Print  only  the  matched  (non-empty) parts of a matching line,
          with each such part on a separate output line.

上下文控制仅限于在匹配之前或之后添加一定数量的行，这在这里可能效果不佳。因此，如果您需要特定字节数的上下文，则必须更改模式本身。

Answer 2

尝试...

grep -rnw "<regex>" <folder>

容易得多。这里有更多示例-> https://computingbro.com/2020/05/10/word-search-in-linux-unix-filesystem/

Grep返回正则表达式导致递归搜索

2 个答案: